Answer and Step-by-step explanation:
The computation is shown below:
As we assume a and b are together and c and d are together
So, the equation would be
[tex]= ab \times cd[/tex]
[tex]= (10a + b) \times (10c + d)[/tex]
= 100ac + 10(ad + bc) + bd
Now another equation i s
[tex]= ba \times cd[/tex]
[tex]= (10b + a) \times (10d + c)[/tex]
= 100bd + 10(ad + bc) + ac
Now take both equations together
So,
100ac + 10(ad + bc) + bd = 100bd + 10(ad + bc) + ac
99ac = 99bd
ac = bd
Now Like here a = 2 b = 10 and c = 5 d = 1
So,
ac = bd = 10
So we can take an example like this
So, by considering the
[tex]62 \times 13 = 806[/tex]
And,
[tex]26 \times 31 = 806[/tex]
We can the same amount
So in the given question, the choices are not there so we can calculate by the above method